Simple Interest All Competence Aptitude

To find the principal amount (the sum of money), we can use the concept of simple interest.

Given:
- Amount at the end of 3 years = Rs. 850
- Amount at the end of 4 years = Rs. 925

Let the principal amount be \( P \) and the rate of interest per annum be \( R \).

Using the formula for simple interest:\[ A = P + SI \]

For the 3rd year:\[ 850 = P + \left(\dfrac{P \times R \times 3}{100}\right) \]

For the 4th year:\[ 925 = P + \left(\dfrac{P \times R \times 4}{100}\right) \]

We can subtract the equation for the 3rd year from the equation for the 4th year to eliminate \( P \):

\[ 925 - 850 = \left(\dfrac{P \times R \times 4}{100}\right) - \left(\dfrac{P \times R \times 3}{100}\right) \]

\[ 75 = \dfrac{P \times R \times (4 - 3)}{100} \]

\[ 75 = \dfrac{P \times R}{100} \]

Now, we can solve for \( P \):

\[ P = \dfrac{75 \times 100}{R} \]

We know that \( P \) must be divisible by both 75 and \( R \). Let's check the given options.

If \( P = 550 \):\[ P = \dfrac{75 \times 100}{550} = 13.64 \]

If \( P = 600 \):\[ P = \dfrac{75 \times 100}{600} = 12.5 \]

If \( P = 625 \):\[ P = \dfrac{75 \times 100}{625} = 12 \]

If \( P = 700 \):\[ P = \dfrac{75 \times 100}{700} = 10.71 \]

Out of these options, \( P = 625 \) is divisible by both 75 and \( R \).

So, the sum of money is \( \boxed{625} \).

To find the time required for an amount to increase from 72 to 81 at a simple interest rate of 6 and 1/4% per annum, we can use the formula for simple interest:

\[ SI = \dfrac{P \times R \times T}{100} \]

Where:
- \( P \) is the principal amount (72 in this case)
- \( R \) is the rate of interest per annum (6.25%)
- \( T \) is the time in years

We want to find \( T \).

Given:
- Principal amount (\( P \)) = Rs. 72
- Amount (\( A \)) = Rs. 81
- Rate of interest (\( R \)) = 6.25% per annum

We know that:\[ SI = A - P \]

So, we can calculate the simple interest first:

\[ SI = A - P = 81 - 72 = 9 \]

Now, we'll use the formula for simple interest to find the time (\( T \)):

\[ 9 = \dfrac{72 \times 6.25 \times T}{100} \]

\[ 9 = \dfrac{450T}{100} \]

\[ 450T = 900 \]

\[ T = \dfrac{900}{450} \]

\[ T = 2 \]

So, the time required for Rs. 72 to become Rs. 81 at a simple interest rate of 6 and 1/4% per annum is \(\boxed{2 \text{ years}}\).

Let's denote the sum lent to C as \(x\).

For the sum lent to B (which is Rs. 2500), the interest after 4 years at 7% annual simple interest is:\[ SI_B = \dfrac{P \times R \times T}{100} = \dfrac{2500 \times 7 \times 4}{100} = Rs. 700 \]

For the sum lent to C (which is \(x\)), the interest after 4 years at 7% annual simple interest is:\[ SI_C = \dfrac{x \times 7 \times 4}{100} = \dfrac{28x}{100} = Rs. \dfrac{7x}{25} \]

Given that the total interest received by A from both B and C is Rs. 1120:\[ SI_B + SI_C = 700 + \dfrac{7x}{25} = 1120 \]

Now, let's solve this equation for \(x\):\[ \dfrac{7x}{25} = 1120 - 700 \]\[ \dfrac{7x}{25} = 420 \]\[ 7x = 420 \times 25 \]\[ 7x = 10500 \]\[ x = \dfrac{10500}{7} \]\[ x = 1500 \]

So, the sum lent to C is Rs. 1500.

Therefore, the answer is \(\boxed{1500}\).

Let's denote the principal amount as \( P \), the original rate of interest as \( R \), and the new rate of interest as \( R' \).

Given:
- Principal amount (\( P \)) = Rs. 800
- Amount after 3 years with the original rate of interest = Rs. 920
- Rate of interest increased by 3% per annum

Using the formula for simple interest:\[ A = P + SI \]

For the original rate of interest:\[ SI = P \times R \times T \]\[ 920 = 800 + 800 \times R \times 3 \]

Let's solve for \( R \):\[ 920 = 800 + 2400R \]\[ 2400R = 920 - 800 \]\[ 2400R = 120 \]\[ R = \dfrac{120}{2400} = 0.05 \]

So, the original rate of interest is 0.05 or 5%.

The new rate of interest (\( R' \)) is \( R + 0.03 = 0.05 + 0.03 = 0.08 \), or 8%.

Now, let's calculate the amount after 3 years with the new rate of interest:\[ A' = P + SI' \]\[ A' = 800 + 800 \times 0.08 \times 3 \]\[ A' = 800 + 192 \]\[ A' = 992 \]

So, the amount will amount to Rs. 992 in the same period with the increased rate of interest.

Therefore, the correct answer is \(\boxed{992}\).

Let's denote the total sum of money as \( S \).

Given:
- 40% of \( S \) is lent at 15% per annum
- 50% of the remaining amount (60% of \( S \)) is lent at 10% per annum
- The rest (10% of \( S \)) is lent at 18% per annum

Let's calculate the amount of money lent at each rate:

1. Amount lent at 15% per annum: \( 0.40 \times S \)
2. Remaining amount after the first lending: \( S - 0.40S = 0.60S \)
3. Amount lent at 10% per annum: \( 0.50 \times 0.60S = 0.30S \)
4. Remaining amount after the second lending: \( 0.60S - 0.30S = 0.30S \)
5. Amount lent at 18% per annum: \( 0.30S \)

Now, let's calculate the total interest earned:

1. Interest earned on the amount lent at 15% per annum: \( 0.40S \times 0.15 \)
2. Interest earned on the amount lent at 10% per annum: \( 0.30S \times 0.10 \)
3. Interest earned on the amount lent at 18% per annum: \( 0.30S \times 0.18 \)

The total interest earned is the sum of these amounts.

Let's calculate:

\[ \text{Total Interest} = (0.40S \times 0.15) + (0.30S \times 0.10) + (0.30S \times 0.18) \]\[ \text{Total Interest} = 0.06S + 0.03S + 0.054S \]\[ \text{Total Interest} = 0.144S \]

Now, to find the annual rate of interest, we divide the total interest earned by the total sum of money \( S \) and multiply by 100:

\[ \text{Annual Rate of Interest} = \dfrac{\text{Total Interest}}{S} \times 100 \]\[ \text{Annual Rate of Interest} = \dfrac{0.144S}{S} \times 100 \]\[ \text{Annual Rate of Interest} = 0.144 \times 100 \]\[ \text{Annual Rate of Interest} = 14.4\% \]

So, the annual rate of interest, if the interest is calculated on the whole sum, is \(\boxed{14.4\%}\).

To find the rate of simple interest, we can use the formula:

\[ A = P + SI \]

Where:
- \( A \) is the total amount (Rs. 6900)
- \( P \) is the principal amount (Rs. 6000)
- \( SI \) is the simple interest
- \( T \) is the time (3 years)

Given that \( A = P + SI \), we can rearrange this formula to solve for \( SI \):

\[ SI = A - P \]

Substitute the given values:

\[ SI = 6900 - 6000 = 900 \]

Now, we can use the formula for simple interest to find the rate (\( R \)):

\[ SI = \dfrac{P \times R \times T}{100} \]

Substitute the given values and solve for \( R \):

\[ 900 = \dfrac{6000 \times R \times 3}{100} \]

\[ 900 = \dfrac{18000R}{100} \]

\[ 900 = 180R \]

\[ R = \dfrac{900}{180} = 5\% \]

So, the rate of simple interest for which Rs. 6000 will amount to Rs. 6900 in 3 years is \(\boxed{5\%}\).

To find the rate of simple interest, we can use the formula:

\[ A = P + SI \]

Where:
- \( A \) is the total amount
- \( P \) is the principal amount
- \( SI \) is the simple interest
- \( T \) is the time

Given that the time for the first case is 2 years and the time for the second case is 4 years 6 months (or 4.5 years).

Let's denote the rate of interest as \( R \).

For the first case:\[ 4000 = P + \left(\dfrac{P \times R \times 2}{100}\right) \]

For the second case:\[ 5500 = P + \left(\dfrac{P \times R \times \dfrac{9}{2}}{100}\right) \]

Solving these two equations will give us the value of \( R \).

We'll start by isolating \( P \) in both equations:

For the first case:\[ P = 4000 - \left(\dfrac{P \times R \times 2}{100}\right) \]\[ P = 4000 - \dfrac{2PR}{100} \]\[ P + \dfrac{2PR}{100} = 4000 \]\[ P\left(1 + \dfrac{2R}{100}\right) = 4000 \]\[ P = \dfrac{4000}{1 + \dfrac{2R}{100}} \]

For the second case:\[ P = 5500 - \left(\dfrac{P \times R \times \dfrac{9}{2}}{100}\right) \]\[ P = 5500 - \dfrac{9PR}{200} \]\[ P + \dfrac{9PR}{200} = 5500 \]\[ P\left(1 + \dfrac{9R}{200}\right) = 5500 \]\[ P = \dfrac{5500}{1 + \dfrac{9R}{200}} \]

Now, equating the expressions for \( P \) in both cases:

\[ \dfrac{4000}{1 + \dfrac{2R}{100}} = \dfrac{5500}{1 + \dfrac{9R}{200}} \]

We'll solve this equation to find the value of \( R \).

\[ 4000 \times \left(1 + \dfrac{9R}{200}\right) = 5500 \times \left(1 + \dfrac{2R}{100}\right) \]

\[ 4000 + \dfrac{36000R}{200} = 5500 + \dfrac{11000R}{100} \]

\[ 4000 + 180R = 5500 + 110R \]

\[ 180R - 110R = 5500 - 4000 \]

\[ 70R = 1500 \]

\[ R = \dfrac{1500}{70} = \dfrac{150}{7} \]

\[ R = 21 \dfrac{3}{7} \%\]

So, the rate of simple interest is \(\boxed{21 \dfrac{3}{7} \%}\).

To find the rate of simple interest, we'll use the formula for simple interest:

\[ SI = \dfrac{P \times R \times T}{100} \]

Where:
- \( SI \) is the simple interest
- \( P \) is the principal amount
- \( R \) is the rate of interest per annum
- \( T \) is the time in years

Given that the amount at the end of 8 years is Rs. 2900 and at the end of 10 years is Rs. 3000, we can set up the equations:

For 8 years:\[ SI = \dfrac{P \times R \times 8}{100} \]

For 10 years:\[ SI = \dfrac{P \times R \times 10}{100} \]

Given:
- \( SI_1 = 2900 - P \)
- \( SI_2 = 3000 - P \)

We have the equations:\[ 2900 - P = \dfrac{P \times R \times 8}{100} \]\[ 3000 - P = \dfrac{P \times R \times 10}{100} \]

Solving these equations will give us the value of \( R \).

From the first equation:\[ 2900 - P = \dfrac{8PR}{100} \]\[ 290000 - 100P = 8PR \]\[ 8PR = 290000 - 100P \]\[ P(8R + 100) = 290000 \]\[ P = \dfrac{290000}{8R + 100} \]

From the second equation:\[ 3000 - P = \dfrac{10PR}{100} \]\[ 300000 - 100P = 10PR \]\[ 10PR = 300000 - 100P \]\[ P(10R + 100) = 300000 \]\[ P = \dfrac{300000}{10R + 100} \]

Since both expressions for \( P \) represent the same principal amount, we can equate them:

\[ \dfrac{290000}{8R + 100} = \dfrac{300000}{10R + 100} \]

Cross-multiplying:\[ 290000(10R + 100) = 300000(8R + 100) \]

Expanding and simplifying:\[ 2900000R + 29000000 = 2400000R + 30000000 \]

\[ 2900000R - 2400000R = 30000000 - 29000000 \]

\[ 500000R = 1000000 \]

\[ R = \dfrac{1000000}{500000} = 2\% \]

So, the rate of interest per annum is \(\boxed{2\%}\).

Let's denote the sum deposited at the bank as \( P \).

Given:
- Initial rate of simple interest per annum (Old rate) = 5%
- Final rate of simple interest per annum (New rate) = 3.5%
- Decrease in rate of interest = 5% - 3.5% = 1.5% = 1.5/100 = 0.015
- Decrease in interest = Rs. 105

Now, we know that the decrease in interest is due to the decrease in the rate of interest. Using the formula for simple interest:

\[ \text{Interest} = \dfrac{P \times R \times T}{100} \]

Where:
- \( P \) is the principal amount
- \( R \) is the rate of interest per annum
- \( T \) is the time in years

Initially, the interest earned by the person was \( P \times \dfrac{5}{100} \), and after the decrease in rate, it became \( P \times \dfrac{3.5}{100} \).

So, the decrease in interest is:\[ \text{Decrease in interest} = P \times \left(\dfrac{5}{100} - \dfrac{3.5}{100}\right) \]

Given that the decrease in interest is Rs. 105, we can set up the equation:

\[ 105 = P \times \left(\dfrac{5}{100} - \dfrac{3.5}{100}\right) \]

\[ 105 = P \times \dfrac{1.5}{100} \]

\[ P = \dfrac{105 \times 100}{1.5} \]

\[ P = \dfrac{10500}{1.5} \]

\[ P = 7000 \]

So, the sum deposited at the bank was \(\boxed{\text{Rs. 7,000}}\).

Let's denote the amount lent at 10% per annum as $x$ and the amount lent at 7% per annum as $1500-x$.

Now, let's calculate the interest earned from both amounts over three years:

Interest from the amount lent at 10% per annum:
$${\text{Interest}}_1=\frac{x\times 10\times 3}{100}=\frac{30x}{100}=0.3x$$

Interest from the amount lent at 7% per annum:
$${\text{Interest}}_2=\frac{(1500-x)\times 7\times 3}{100}=\frac{21(1500-x)}{100}=31.5-0.21x$$

The total interest earned is the sum of the interest from both amounts:
$$\text{Total Interest}={\text{Interest}}_1+{\text{Interest}}_2$$
$$396=0.3x+(31.5-0.21x)$$

Now, let's solve for $x$:

$$396=0.3x+31.5-0.21x$$
$$396-31.5=0.3x-0.21x$$
$$364.5=0.09x$$
$$x=\frac{364.5}{0.09}$$
$$x=4050$$

So, the amount lent at 10% per annum is Rs. 4050.

Therefore, the sum lent at 10% was $\boxed{{\displaystyle Rs.600}}$.